Let $X$ be a smooth projective and geometrically irreducible curve over a field $k$. It an exercise in Hartshorne's book on "Algebraic Geometry" (p.149) that in the case that $k$ is algebraically closed the Grothendieck group $K(X)$ is isomorphic to $\textrm{Pic}(X)\oplus\mathbb{Z}$ (and the isomorphism is given by determinant and rank). Is this also the case when $X$ is not algebraically closed? It seems to me on the first glimpse that the same proof would work but then I am a bit insecure because why does Hartshorne then not state it for arbitrary fields?
Grothendieck group of a smooth curve
algebraic-curvesalgebraic-geometrysheaf-theory
Related Solutions
Unfortunately, your requested statements are not correct as written. The equivalence of categories can be fixed, but the "in other words" statement is false. Fortunately, there is a relatively quick and easy resolution to the question you ask at the end: any smooth variety which isn't geometrically irreducible splits over some finite extension in to a union of smooth geometrically irreducible varieties. So as long as you don't mind altering your base field up to a finite extension and you talk about the right sort of properties, you don't lose anything by assuming geometrically irreducible.
Let's handle the "in other words" statement first. If $X$ is smooth and birational to a geometrically irreducible variety, then $X$ must be geometrically irreducible, which is not exactly what you've written down - any smooth variety which is not geometrically irreducible is a counterexample to your statement. There are plenty such varieties: $\operatorname{Spec} \Bbb F_2[x]/(x^2+x+1)$, for instance, is smooth but not geometrically irreducible as over $\Bbb F_4$, it splits in to two smooth points.
To show the claim that smooth + birational to a geometrically irreducible variety implies geometrically irreducible, let $X$ be irreducible and smooth over $k$. First we note that any open subscheme of a geometrically irreducible subscheme is again geometrically irreducible, and geometric irreduciblity is preserved by isomorphisms. So by birationality, this means that there is an open subscheme $U$ of $X$ which is geometrically irreducible (ie $U_\overline{k}$ is irreducible). On the other hand, as $X$ is smooth over $k$, $X_\overline{k}$ is smooth over $\overline{k}$, so it's a disjoint union of irreducible components, each of which surjects on to $X$. In particular, the preimage of the generic point of $X$ is the collection of generic points of each irreducible component of $X_\overline{k}$. On the other hand, as $U$ is geometrically irreducible and has the same generic point as $X$, there can only be one of these generic points, so $X$ is geometrically irreducible.
Now let's get to the statement about categories. The easiest way to fix this up is to imitate one of the well-known equivalences between certain categories of curves over $k$ and fg transcendence degree one field extensions of $k$ and then add the condition about a algebraic closure of $k$ in this field from the comments to guarantee geometric irreducibility (the comments reference Liu's AGAC 3.2.14). For a reference on the equivalence of categories, see for instance Hartshorne I.6.12 or Stacks 0BY1. For instance, one possible correct statement would be that there's an equivalence of categories between smooth proper geometrically irreducible curves over $k$ and finitely generated field extensions $k\subset K$ of transcendence degree one with $k$ algebraically closed in $K$.
You need to somehow connect your two resolutions $0\to \mathcal{E}_1\to \mathcal{E}_0\to \mathcal{F}\to 0$ and $0\to \mathcal{E}_1'\to \mathcal{E}_0'\to \mathcal{F}\to 0$. What if you considered $\mathcal{E}_0\oplus\mathcal{E}_0'\to \mathcal{F}$? Can you make any nice-looking homological algebra diagrams from this? You will need some homological algebra lemmas here, and they might not be the most commonly used ones.
You're on the right track by starting with an isomorphism $\mathcal{F}_\xi\cong \mathcal{O}_{X,\xi}^n$. You need to improve this twice: first, promote this to an isomorphism $\mathcal{O}_X^n|_U\to\mathcal{F}|_U$ for some open $U$; second, obtain an injection on the whole curve which after some fiddling will give you the first map in your exact sequence. Once you have the map in your exact sequence, write $\mathcal{T}$ in another way and play around with arithmetic in the Grothendieck group (you may find (a) helpful here).
Best Answer
Yes, the proof from exercise II.6.11 will show that for a regular projective curve $X$ over any field $k$ the Grothendieck group is $\operatorname{Pic}(X)\oplus\Bbb Z$ (regular is weaker than smooth and there is no need to add a geometrically irreducible hypothesis). One slight adjustment to the strategy as outlined by Hartshorne is over a non-algebraically closed field one may have to be a touch more careful with the skyscraper sheaf section of the argument - instead of saying
one should amend this statement to $k(P_i)$ being the skyscraper sheaf with value the residue field of $P_i$ at $P_i$ and 0 elsewhere.
As to why Hartshorne would state things this way, Hartshorne spends a lot of time over algebraically closed fields, especially when doing more concrete geometry, and doesn't always seem to care so much about the non-algebraically-closed case. You'd have to ask him why, but it's possible that's just where his head was 50 years ago when he wrote the text.
(Let me also point out that the result holds in much greater generality: the result is true for any separated noetherian normal connected scheme of dimension one via essentially the same proof - some improved argument is needed to justify that such a scheme has the resolution property, but it's not such an obscure proof.)